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Evaluation of $\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$

Now Let $\displaystyle \tan x = t^2\;,$ Then $\sec^2 xdx = 2tdt$ or $\displaystyle dx = \frac{2t}{1+t^4}dt$

So Integral is $\displaystyle \int\frac{t}{1+t}\cdot \frac{2t}{1+t^4}dt = 2\int\frac{t^2}{(1+t)\cdot (1+t^4)}dt$

Now How can I solve after that

Help me

Thanks

Martin Sleziak
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juantheron
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2 Answers2

9

By collecting all the suggestions, you should be able to prove that: $$\frac{2t}{(t+1)(t^4+1)}=\frac{t}{1+t^4}+\frac{t^3}{1+t^4}+\frac{1-t^2}{1+t^4}-\frac{1}{1+t},$$ and since $(1+t^4)=(1+\sqrt{2}t+t^2)(1-\sqrt{2}t+t^2)$, it follows that: $$\int\frac{2t\,dt}{(t+1)(t^4+1)}=\frac{1}{2}\arctan t^2+\frac{1}{4}\log(1+t^4)+\frac{1}{2\sqrt{2}}\log\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}-\log(1+t).$$ $\\$

Addendum

Suggestion for evaluating $\displaystyle{\int \frac{1 - t^2}{t^4 + 1}\,\mathrm{d}t}$: Write $$ \begin{aligned} \int- \frac{t^2 - 1}{t^4 + 1}\,\mathrm{d}t &= -\int \frac{1 - \dfrac{1}{t^2}}{t^2 + \dfrac{1}{t^2}}\,\mathrm{d}t \\ &=-\int \frac{1 - \dfrac{1}{t^2}}{\left(t + \dfrac{1}{t}\right)^2 - 2}\,\mathrm{d}t \end{aligned} $$

Now, set $\displaystyle{u = t + \frac{1}{t}}$ and $\mathrm{d}u = \left(1 - \dfrac{1}{t^2}\right)\,\mathrm{d}t$: $$ \begin{aligned} -\int \frac{\mathrm{d}u}{u^2 - 2} &= \dfrac{1}{2\sqrt{2}}\ln\left|\frac{u + \sqrt{2}}{u - \sqrt{2}} \right| + C \\ &=\frac{1}{2\sqrt{2}}\ln\left|\frac{t^2 + t\sqrt{2} +1}{t^2 - t\sqrt{2}+1} \right| + C \end{aligned} $$ Note that $t^2 + t\sqrt{2}+1 = \left(t + \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$ and $t^2 - t\sqrt{2}+1 = \left(t - \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$, then we can get rid of the absolute value bars: $$\int \frac{1-t^2}{t^4 + 1}\,\mathrm{d}t = \frac{1}{2\sqrt{2}}\ln\!\left(\dfrac{t^2 + t\sqrt{2}+1}{t^2 - t\sqrt{2}+1}\right) + C $$

This is a very famous technique.

user149844
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Jack D'Aurizio
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2

$\int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\int\frac{\sqrt{\sin x}(\sqrt{\sin x}-\sqrt{\cos x})}{\sin x-\cos x}dx=\int\frac{\sin x}{\sin x-\cos x}dx-\frac{1}{\sqrt 2}\int\frac{\sqrt{\sin 2x}}{\sin x-\cos x}dx$

Let $I_1=\int\frac{\sin x}{\sin x-\cos x}dx=\frac{1}{2}\int\frac{\sin x-\cos x+(\cos x+\sin x)}{\sin x-\cos x}dx=\frac{x}{2}+\frac{1}{2}ln|\sin x-\cos x|+c$

$$\int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\frac{x}{2}+\frac{1}{2}ln|\sin x-\cos x|-\frac{1}{\sqrt 2}\int\frac{\sqrt{\sin 2x}}{\sin x-\cos x}dx$$

Let $I=\int\frac{\sqrt{\sin 2x}}{\sin x-\cos x}dx=\int\frac{\sqrt{\sin 2x}(\sin x-\cos x)}{1-\sin 2x}dx=\int\frac{\sqrt{(\sin x+\cos x)^2-1}}{2-(\sin x+\cos x)^2}(\sin x-\cos x)dx$

Put $t=\sin x+\cos x$

$I=\int\frac{\sqrt{t^2-1}}{2-t^2}dt$

Put $t=\sec u$

$I=\int\frac{\tan u}{2-\sec^2 u}(\sec u\tan u)du=\int\frac{\sin^2 u}{\cos u(2\cos^2 u-1)}du=\int\frac{\sin^2 u}{(1-\sin^2 u)(1-2\sin^2 u)}(\cos u)du$

$I=\int\left(\frac{1}{1-2\sin^2u}-\frac{1}{1-\sin^2 u}\right)(\cos u)du=-\frac{1}{2\sqrt2}ln|\frac{\sqrt2\sin u-1}{\sqrt2\sin u+1}|+\frac{1}{2}ln|\frac{\sin u-1}{\sin u+1}|+c$

$I=-\frac{1}{2\sqrt2}ln|\frac{\sin x+\cos x-\sqrt{2\sin 2x}}{\sin x+\cos x+\sqrt{2\sin 2x}}|+\frac{1}{2}ln|\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}|+c$

$$\int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\frac{x}{2}+\frac{1}{2}ln|\sin x-\cos x|-\frac{1}{4}ln|\frac{\sin x+\cos x-\sqrt{2\sin 2x}}{\sin x+\cos x+\sqrt{2\sin 2x}}|+\frac{1}{2\sqrt2}ln|\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}|+c$$

Maverick
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